\(\int \frac {(a g+b g x)^2}{A+B \log (\frac {e (a+b x)^2}{(c+d x)^2})} \, dx\) [137]

Optimal result

Integrand size = 34, antiderivative size = 34 \[ \int \frac {(a g+b g x)^2}{A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )} \, dx=\text {Int}\left (\frac {(a g+b g x)^2}{A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )},x\right ) \]

[Out]

Unintegrable((b*g*x+a*g)^2/(A+B*ln(e*(b*x+a)^2/(d*x+c)^2)),x)

Rubi [N/A]

Not integrable

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {(a g+b g x)^2}{A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )} \, dx=\int \frac {(a g+b g x)^2}{A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )} \, dx \]

[In]

Int[(a*g + b*g*x)^2/(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]),x]

[Out]

Defer[Int][(a*g + b*g*x)^2/(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {(a g+b g x)^2}{A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )} \, dx \\ \end{align*}

Mathematica [N/A]

Not integrable

Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {(a g+b g x)^2}{A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )} \, dx=\int \frac {(a g+b g x)^2}{A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )} \, dx \]

[In]

Integrate[(a*g + b*g*x)^2/(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]),x]

[Out]

Integrate[(a*g + b*g*x)^2/(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]), x]

Maple [N/A]

Not integrable

Time = 0.72 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00

\[\int \frac {\left (b g x +a g \right )^{2}}{A +B \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right )}d x\]

[In]

int((b*g*x+a*g)^2/(A+B*ln(e*(b*x+a)^2/(d*x+c)^2)),x)

[Out]

int((b*g*x+a*g)^2/(A+B*ln(e*(b*x+a)^2/(d*x+c)^2)),x)

Fricas [N/A]

Not integrable

Time = 0.25 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.21 \[ \int \frac {(a g+b g x)^2}{A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )} \, dx=\int { \frac {{\left (b g x + a g\right )}^{2}}{B \log \left (\frac {{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + A} \,d x } \]

[In]

integrate((b*g*x+a*g)^2/(A+B*log(e*(b*x+a)^2/(d*x+c)^2)),x, algorithm="fricas")

[Out]

integral((b^2*g^2*x^2 + 2*a*b*g^2*x + a^2*g^2)/(B*log((b^2*e*x^2 + 2*a*b*e*x + a^2*e)/(d^2*x^2 + 2*c*d*x + c^2
)) + A), x)

Sympy [N/A]

Not integrable

Time = 14.06 (sec) , antiderivative size = 258, normalized size of antiderivative = 7.59 \[ \int \frac {(a g+b g x)^2}{A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )} \, dx=g^{2} \left (\int \frac {a^{2}}{A + B \log {\left (\frac {a^{2} e}{c^{2} + 2 c d x + d^{2} x^{2}} + \frac {2 a b e x}{c^{2} + 2 c d x + d^{2} x^{2}} + \frac {b^{2} e x^{2}}{c^{2} + 2 c d x + d^{2} x^{2}} \right )}}\, dx + \int \frac {b^{2} x^{2}}{A + B \log {\left (\frac {a^{2} e}{c^{2} + 2 c d x + d^{2} x^{2}} + \frac {2 a b e x}{c^{2} + 2 c d x + d^{2} x^{2}} + \frac {b^{2} e x^{2}}{c^{2} + 2 c d x + d^{2} x^{2}} \right )}}\, dx + \int \frac {2 a b x}{A + B \log {\left (\frac {a^{2} e}{c^{2} + 2 c d x + d^{2} x^{2}} + \frac {2 a b e x}{c^{2} + 2 c d x + d^{2} x^{2}} + \frac {b^{2} e x^{2}}{c^{2} + 2 c d x + d^{2} x^{2}} \right )}}\, dx\right ) \]

[In]

integrate((b*g*x+a*g)**2/(A+B*ln(e*(b*x+a)**2/(d*x+c)**2)),x)

[Out]

g**2*(Integral(a**2/(A + B*log(a**2*e/(c**2 + 2*c*d*x + d**2*x**2) + 2*a*b*e*x/(c**2 + 2*c*d*x + d**2*x**2) +
b**2*e*x**2/(c**2 + 2*c*d*x + d**2*x**2))), x) + Integral(b**2*x**2/(A + B*log(a**2*e/(c**2 + 2*c*d*x + d**2*x
**2) + 2*a*b*e*x/(c**2 + 2*c*d*x + d**2*x**2) + b**2*e*x**2/(c**2 + 2*c*d*x + d**2*x**2))), x) + Integral(2*a*
b*x/(A + B*log(a**2*e/(c**2 + 2*c*d*x + d**2*x**2) + 2*a*b*e*x/(c**2 + 2*c*d*x + d**2*x**2) + b**2*e*x**2/(c**
2 + 2*c*d*x + d**2*x**2))), x))

Maxima [N/A]

Not integrable

Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {(a g+b g x)^2}{A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )} \, dx=\int { \frac {{\left (b g x + a g\right )}^{2}}{B \log \left (\frac {{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + A} \,d x } \]

[In]

integrate((b*g*x+a*g)^2/(A+B*log(e*(b*x+a)^2/(d*x+c)^2)),x, algorithm="maxima")

[Out]

integrate((b*g*x + a*g)^2/(B*log((b*x + a)^2*e/(d*x + c)^2) + A), x)

Giac [N/A]

Not integrable

Time = 0.53 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {(a g+b g x)^2}{A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )} \, dx=\int { \frac {{\left (b g x + a g\right )}^{2}}{B \log \left (\frac {{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + A} \,d x } \]

[In]

integrate((b*g*x+a*g)^2/(A+B*log(e*(b*x+a)^2/(d*x+c)^2)),x, algorithm="giac")

[Out]

integrate((b*g*x + a*g)^2/(B*log((b*x + a)^2*e/(d*x + c)^2) + A), x)

Mupad [N/A]

Not integrable

Time = 2.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {(a g+b g x)^2}{A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )} \, dx=\int \frac {{\left (a\,g+b\,g\,x\right )}^2}{A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^2}\right )} \,d x \]

[In]

int((a*g + b*g*x)^2/(A + B*log((e*(a + b*x)^2)/(c + d*x)^2)),x)

[Out]

int((a*g + b*g*x)^2/(A + B*log((e*(a + b*x)^2)/(c + d*x)^2)), x)